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SPM Additional Mathematics 2023, Paper 1 (Question 14)


Question 14:
(a) It is given that $$ f(x)=2 x-5 \text { and } g f(x)=\frac{4}{2 x-5}, x \neq h $$
State the value of h. [1 mark]

(b) Find the value of x when f(x) maps onto itself.  [2 marks]

(c) Find
(i)  g(x),
(ii) g(4x-3)(x).
[5 marks]

Solution:
(a)
$$ \begin{aligned} g f(x) & =\frac{4}{2 x-5} \\ 2 x-5 & \neq 0 \\ 2 x & \neq 5 \\ x & \neq \frac{5}{2} \\ \therefore h & =\frac{5}{2} \end{aligned} $$

(b)
$$ \begin{aligned} f(x) & =x \\ 2 x-5 & =x \\ x & =5 \end{aligned} $$

(c)(i)
$$ \begin{aligned} g f(x) & =\frac{4}{2 x-5} \\ g(2 x-5) & =\frac{4}{2 x-5} \\ \text { Let } 2 x-5 & =m \\ 2 x & =m+5 \\ x & =\frac{m+5}{2} \end{aligned} $$
$$ \begin{aligned} g(m) & =\frac{4}{2\left(\frac{m+5}{2}\right)-5} \\ & =\frac{4}{m+5-5} \\ & =\frac{4}{m} \\ \therefore g(x) & =\frac{4}{x}, x \neq 0 \end{aligned} $$
(c)(ii)
$$ \begin{aligned} g(x) & =\frac{4}{x} \\ g^2(x) & =g\left(\frac{4}{x}\right) \\ & =\frac{4}{\frac{4}{x}} \\ & =x \end{aligned} $$
$$ \begin{aligned} g^3(x) & =g g^2(x) \\ & =g(x) \\ g^3(x) & =\frac{4}{x} \end{aligned} $$

$$ \begin{aligned} g^4(x) & =g^2 g^2(x) \\ & =g^2(x) \\ g^4(x) & =x \end{aligned} $$
$$ \begin{aligned} &\begin{aligned} & g^{\text {odd }}(x)=\frac{4}{x} \\ & g^{\text {even }}(x)=x \end{aligned}\\ &4 n-3 \text { is odd for all integer, } n \text {. }\\ &\therefore g^{4 n-3}(x)=\frac{4}{x}, x \neq 0 \end{aligned} $$

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